Solve x^2+1/x+2+2x+4/x^2+1=3

$$\frac { x ^ { 2 } + 1 } { x + 2 } + \frac { 2 x + 4 } { x ^ { 2 } + 1 } = 3$$

Solve for x

$$x=\frac{1-\sqrt{5}}{2}\approx -0.618033989$$

$$x=3$$

$$x = \frac{\sqrt{5} + 1}{2} \approx 1.618033989$$

$$x=-1$$

Solution Steps

Variable x cannot be equal to -2 since division by zero is not defined. Multiply both sides of the equation by \left(x+2\right)\left(x^{2}+1\right), the least common multiple of x+2,x^{2}+1.

$$\left(x^{2}+1\right)\left(x^{2}+1\right)+\left(x+2\right)\left(2x+4\right)=3\left(x+2\right)\left(x^{2}+1\right)$$

Multiply x^{2}+1 and x^{2}+1 to get \left(x^{2}+1\right)^{2}.

$$\left(x^{2}+1\right)^{2}+\left(x+2\right)\left(2x+4\right)=3\left(x+2\right)\left(x^{2}+1\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x^{2}+1\right)^{2}.

$$\left(x^{2}\right)^{2}+2x^{2}+1+\left(x+2\right)\left(2x+4\right)=3\left(x+2\right)\left(x^{2}+1\right)$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$x^{4}+2x^{2}+1+\left(x+2\right)\left(2x+4\right)=3\left(x+2\right)\left(x^{2}+1\right)$$

Use the distributive property to multiply x+2 by 2x+4 and combine like terms.

$$x^{4}+2x^{2}+1+2x^{2}+8x+8=3\left(x+2\right)\left(x^{2}+1\right)$$

Combine 2x^{2} and 2x^{2} to get 4x^{2}.

$$x^{4}+4x^{2}+1+8x+8=3\left(x+2\right)\left(x^{2}+1\right)$$

Add 1 and 8 to get 9.

$$x^{4}+4x^{2}+9+8x=3\left(x+2\right)\left(x^{2}+1\right)$$

Use the distributive property to multiply 3 by x+2.

$$x^{4}+4x^{2}+9+8x=\left(3x+6\right)\left(x^{2}+1\right)$$

Use the distributive property to multiply 3x+6 by x^{2}+1.

$$x^{4}+4x^{2}+9+8x=3x^{3}+3x+6x^{2}+6$$

Subtract 3x^{3} from both sides.

$$x^{4}+4x^{2}+9+8x-3x^{3}=3x+6x^{2}+6$$

Subtract 3x from both sides.

$$x^{4}+4x^{2}+9+8x-3x^{3}-3x=6x^{2}+6$$

Combine 8x and -3x to get 5x.

$$x^{4}+4x^{2}+9+5x-3x^{3}=6x^{2}+6$$

Subtract 6x^{2} from both sides.

$$x^{4}+4x^{2}+9+5x-3x^{3}-6x^{2}=6$$

Combine 4x^{2} and -6x^{2} to get -2x^{2}.

$$x^{4}-2x^{2}+9+5x-3x^{3}=6$$

Subtract 6 from both sides.

$$x^{4}-2x^{2}+9+5x-3x^{3}-6=0$$

Subtract 6 from 9 to get 3.

$$x^{4}-2x^{2}+3+5x-3x^{3}=0$$

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{4}-3x^{3}-2x^{2}+5x+3=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 3 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±3,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}-3x^{3}-2x^{2}+5x+3 by x+1 to get x^{3}-4x^{2}+2x+3. Solve the equation where the result equals to 0.

$$x^{3}-4x^{2}+2x+3=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 3 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±3,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-4x^{2}+2x+3 by x-3 to get x^{2}-x-1. Solve the equation where the result equals to 0.

$$x^{2}-x-1=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -1 for b, and -1 for c in the quadratic formula.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 1\left(-1\right)}}{2}$$

Do the calculations.

$$x=\frac{1±\sqrt{5}}{2}$$

Solve the equation x^{2}-x-1=0 when ± is plus and when ± is minus.

$$x=\frac{1-\sqrt{5}}{2} x=\frac{\sqrt{5}+1}{2}$$

List all found solutions.

$$x=-1 x=3 x=\frac{1-\sqrt{5}}{2} x=\frac{\sqrt{5}+1}{2}$$